Accurate Measurements on pulsed system's harder than you think.

[EMJunkie]

I agree that Laurent made a great clip.  The one missing thing is one simple thing:  You can check it by working it out on paper, you don't need to rely on any derived meter readings at all, just the basic measurements will suffice.  The frequency is low so we can ignore reactive effects.

Let's say the current is when the MOSFET is ON is (12.22 volts/101 ohms) = 121 milliamps.

Let's say the voltage is (12.22 volts x 100/101) = 12.10 volts

The instantaneous power is 12.10 volts x 0.121 amps = 1.46 watts

Let's keep it simple and say the 20% duty cycle is ON for one second and OFF for four seconds.  So that 1.46 Joules for every five seconds.

The average power is (1.46 Joules/5 seconds) = 0.292 watts of average power.  We calculated this ourselves and know this to be true within the limits of the data we collected.

If we use the scope voltage of 1.26 volts we calculate the average power as 0.315 watts.

If Laurent scoped the voltage output from the battery and took accurate measurements of the resistances of the two resistors and did as accurate a measurement as possible of the duty cycle with his scope we could fine tune our calculation.  The voltage output from the battery almost certainly takes a dip when the MOSFET switches on and that is an unknown.  We don't know what the voltage drop across the MOSFET is.

The scope method using peak values and factoring in the duty cycle measured 0.337 watts.
The analog meter method measured 0.0875 watts.
The scope averaging method measured 0.0559 watts.

It looks like the scope method using peak values and factoring in the duty cycle is the "correct" value.  I put correct in quotations because your calculated values and your measured values have to be much closer together than we see to get satisfaction.  Why are you seeing these discrepancies?  If you are serious it merits a serious follow-up investigation.  How can you be sure if any of your measurements are going to be good if you don't understand exactly what is taking place in this clip?

There are two things to contemplate here that can affect what's going on:

1.  This is not truly a circuit that pulses a voltage (i.e. the drive voltage for the circuit) that goes from zero volts to 12 volts.  What it is is a MOSFET switch that makes a connection and then breaks a connection and goes high-impedance.  There is no "zero volt" signal.  Open-circuit can be way way different from zero volts, especially as you go towards higher frequencies.

2.  It would be very interesting to see what you would measure with digital multimeters for the voltage and the current measurements.  Do not use an in-line ammeter, just use a digital multimeter to measure the average voltage across the one-ohm resistor.  Digital multimeters can be deadly accurate at making average measurements because they are integration-based measuring devices.

It's amazing how the simple measurements in this simple circuit do indeed present a challenge to the experimenter.  If I was in your shoes, I would investigate this from A to Z and understand exactly what was happening.  Do not forget to use your own head, you don't necessarily need to use the full measurement capabilities of your instruments.  Just do the basic measurements and then do the calculations on paper.  That can be used to double-check what is going on.

MileHigh

Great Video, thanks for sharing Woopy!

I am very impressed MileHigh, very helpful, constructive post! This is the stuff that really does help people advance in their understanding to better see what it is that they are looking at.

Very nice!!!

   Chris Sykes
       hyiq.org

[tinman]

A closer look at Woopy's test result's going on the scope shot below.

Calculating by instantaneous power measurements ,and then dividing by 5-as Woopy has a 20% duty cycle.
Version A
Vcc on channel 1 is 12.6v
Vcc on channel 2 is 134mV
CVR is 1 ohm= 134mA
12.6 x 134mA= 1.688 watts / 5 =337.6mW average
Power dissipated by CVR is-->134mV / 5=26.8mV over 1 ohm= .718mW

Version B
Calculating by mean value current.
Vcc on channel 1 is 12.6
V mean channel 2 is 20.4mV over 1 ohm= an average current of 20.4mA
12.6 x 20.4mA = 257.04mW average
Power dissipated by CVR is .416mW

Version C-analog meters.

Amp's = 35mA
Volt's= 2.5 volt's plus the average voltage drop across the CVR of 20mV = 2.52v
2.52v X 35mA = 088mW

Version D
A look at the 100 ohm load resistor.
Scope says Vcc of 12.6v.
12.6v over 100 ohms give us an instantaneous power of 1.587 watts
Divide by 5 for 20% duty cycle,we get an average power dissipation of 317mW
The average current is there for 25.2mA

Version E
Calculating using both mean value's.
2.74v X 24.04mA = 65.87mW.


Version A says our average current should be 26.8mA
Version B (Scope mean value) says average current should be 20.4mA
Analog meter says an average current of 35mA
100 ohm resistor is saying the average current should be 25.2mA

The results show by Woopy's test are as follows

Version A = 337.6mW
Version B = 257.04mW
Version C = 88mW
Version D =317mW <--CVR power dissipation not accounted for here.
Version E =65.87mW

Not one answer the same.
So which one is correct?
The only way to find out is to do a load resistor heat dissipation test with an applied DC current to the load resistor in the DUT,and then compare results against the calculated P/in of that DUT.


Brad

[TinselKoala]

You are STILL doing it! Leading yourselves "up the creek" by following your wild Red Herrings.

The meters used by Woopy are giving the approximate _average values_ of the voltage and current he is measuring. We know this to be true because 1) Poynt99 has done considerable research and testing to confirm that DMMs are good at this; and 2) The values agree within the error range with what the Scope is calculating. (Don't forget that the Analog ammeter is not even calibrated and cannot be read precisely anyhow.)  Multiplying these _average_ values to get some kind of figure and calling that "average Power" in Watts is INCORRECT.

Please review my long, numerical post up above where I take you through the calculations that prove this.

Of course "Accurate Measurements on pulsed system's harder than you think." ... when you use the _incorrect_ procedures for doing the measurement, you use uncalibrated instruments and non-precision sense resistors, and you let yourself chase after Red Herrings. They are actually straightforward, if not exactly "easy", when you use proper instruments, proper procedures and proper sensors, and you have an understanding of the errors and where they come from.

1. The wirewound power resistors Woopy used have a tolerance. They are not guaranteed to be exactly 1.0 ohms or 100.0 ohms. Do they have a "J" printed on them next to the value? This means they could vary by +/- 5 percent from the printed value. Do they have a "K" printed on them? This means +/- 10 percent from the printed value. If you think your resistor is 1.0 ohms but it is actually 0.95 ohms, or even 0.90 ohms, or 1.10 ohms, how does this affect your current _reading_ ?

2. Meters, whether they are analog or digital, have an _insertion loss_ created when they are put into a series circuit as ammeters. In addition, they automatically _average_ the values of pulsed signals, as Poynt99 has shown and as anyone who has experience with moving-coil meters knows. In even more addition, the moving coil ammeter has an unknown calibration and is impossible to read to the precision necessary for this measurement.  For this measurement the "average values" from the meters are useless, because as I have demonstrated, using them basically puts the duty cycle into the calculation twice when it should only be used once. (There are other power measurements, like the total input power of the circuit, where meters can be used properly, but that is not the measurement under examination here.)

3. The Duty Cycle of the pulsations is critical when you are trying to do this calculation by hand rather than letting the scope do the instantaneous multiplication and _subsequent_ averaging to get an average Power value. If the Duty Cycle is actually 22 percent instead of 20 percent you are using, what effect does this have on your "average" calculation? Can you see this much difference by eyeballing the trace on the screen?

4. Even the oscilloscope's accuracy can be questioned. Is there a baseline offset that carries through all the measurements, affecting accuracy? Have you adjusted your calculations (and the scope's calculations) to account for any offset your instrument might have? How precise is an 8-bit analog-to-digital converter when measuring voltages to show on the scope's screen?

5. When trying to analyze a circuit you should include everything in your schematic, and you shouldn't leave scope reference leads dangling. In this case Woopy's dangling reference clip is connected, through all the scope's probe cabling shields and chassis, to the same place the other probe's reference clip is connected. And the pulser circuit is also connected to both the mosfet Gate and to the circuit's negative rail (mosfet Source). At very low frequencies these facts may not matter much, but as you increase the operating frequency they will matter more and more, and sloppy technique should not be practiced or it will become habitual.


In short.... averaging before multiplying V and I does NOT yield a valid " average power " result, especially if you are using analog meters. And any result you get from any calculations, done by hand or by machine or by instrument, is subject to the errors in the input values. Know how these errors affect your results!  Components have tolerances, meters have offsets, ADCs have limits in precision. Know your instruments, and know how they actually work when calculating things like instantaneous multiplication of traces, signal average values, etc.

[TinselKoala]

A closer look at Woopy's test result's going on the scope shot below.

Calculating by instantaneous power measurements ,and then dividing by 5-as Woopy has a 20% duty cycle.
Version A
Vcc on channel 1 is 12.6v
Vcc on channel 2 is 134mV
CVR is 1 ohm= 134mA
12.6 x 134mA= 1.688 watts / 5 =337.6mW average
Power dissipated by CVR is-->134mV / 5=26.8mV over 1 ohm= .718mW

Let's use the correct terms and values and see what is actually happening here.

In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the V_drop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the V_drop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.

Let us "assume" that the 1 Ohm resistor is accurate, though.

So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the V_drop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?

R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.

The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I²R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.

The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I²R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.

A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)

Version B
Calculating by mean value current.
Vcc on channel 1 is 12.6
V mean channel 2 is 20.4mV over 1 ohm= an average current of 20.4mA
12.6 x 20.4mA = 257.04mW average
Power dissipated by CVR is .416mW

The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.

(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)

Why are you multiplying the V_drop across the 100R by the "average current"?  What does this value mean? The power dissipated in the 100 R resistor is calculated by I²R, or equivalently V²/R.

This calculation you've made here is totally invalid.

Version C-analog meters.

Amp's = 35mA
Volt's= 2.5 volt's plus the average voltage drop across the CVR of 20mV = 2.52v
2.52v X 35mA = 088mW

Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.

Version D
A look at the 100 ohm load resistor.
Scope says Vcc of 12.6v.
12.6v over 100 ohms give us an instantaneous power of 1.587 watts
Divide by 5 for 20% duty cycle,we get an average power dissipation of 317mW
The average current is there for 25.2mA

Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V²/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.

Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)²/94.3 =  1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.

Version E
Calculating using both mean value's.
2.74v X 24.04mA = 65.87mW.

Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.

Version A says our average current should be 26.8mA
Version B (Scope mean value) says average current should be 20.4mA
Analog meter says an average current of 35mA
100 ohm resistor is saying the average current should be 25.2mA

The results show by Woopy's test are as follows

Version A = 337.6mW
Version B = 257.04mW
Version C = 88mW
Version D =317mW <--CVR power dissipation not accounted for here.
Version E =65.87mW

Not one answer the same.
So which one is correct?
The only way to find out is to do a load resistor heat dissipation test with an applied DC current to the load resistor in the DUT,and then compare results against the calculated P/in of that DUT.


Brad

You aren't doing it right !!!!!!!!

In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!

In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope.  This is a significant error you have introduced by using the wrong value for the duty cycle.

In the third place you are going badly wrong in some of your calculations.

The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):

The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.

Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.

[EMJunkie]

Thank you TK, two very detailed Posts there! This is the stuff that needs to be saved and referenced at times of taking measurements.

You are STILL doing it! Leading yourselves "up the creek" by following your wild Red Herrings.

The meters used by Woopy are giving the approximate _average values_ of the voltage and current he is measuring. We know this to be true because 1) Poynt99 has done considerable research and testing to confirm that DMMs are good at this; and 2) The values agree within the error range with what the Scope is calculating. (Don't forget that the Analog ammeter is not even calibrated and cannot be read precisely anyhow.)  Multiplying these _average_ values to get some kind of figure and calling that "average Power" in Watts is INCORRECT.

Please review my long, numerical post up above where I take you through the calculations that prove this.

Of course "Accurate Measurements on pulsed system's harder than you think." ... when you use the _incorrect_ procedures for doing the measurement, you use uncalibrated instruments and non-precision sense resistors, and you let yourself chase after Red Herrings. They are actually straightforward, if not exactly "easy", when you use proper instruments, proper procedures and proper sensors, and you have an understanding of the errors and where they come from.

1. The wirewound power resistors Woopy used have a tolerance. They are not guaranteed to be exactly 1.0 ohms or 100.0 ohms. Do they have a "J" printed on them next to the value? This means they could vary by +/- 5 percent from the printed value. Do they have a "K" printed on them? This means +/- 10 percent from the printed value. If you think your resistor is 1.0 ohms but it is actually 0.95 ohms, or even 0.90 ohms, or 1.10 ohms, how does this affect your current _reading_ ?

2. Meters, whether they are analog or digital, have an _insertion loss_ created when they are put into a series circuit as ammeters. In addition, they automatically _average_ the values of pulsed signals, as Poynt99 has shown and as anyone who has experience with moving-coil meters knows. In even more addition, the moving coil ammeter has an unknown calibration and is impossible to read to the precision necessary for this measurement.  For this measurement the "average values" from the meters are useless, because as I have demonstrated, using them basically puts the duty cycle into the calculation twice when it should only be used once. (There are other power measurements, like the total input power of the circuit, where meters can be used properly, but that is not the measurement under examination here.)

3. The Duty Cycle of the pulsations is critical when you are trying to do this calculation by hand rather than letting the scope do the instantaneous multiplication and _subsequent_ averaging to get an average Power value. If the Duty Cycle is actually 22 percent instead of 20 percent you are using, what effect does this have on your "average" calculation? Can you see this much difference by eyeballing the trace on the screen?

4. Even the oscilloscope's accuracy can be questioned. Is there a baseline offset that carries through all the measurements, affecting accuracy? Have you adjusted your calculations (and the scope's calculations) to account for any offset your instrument might have? How precise is an 8-bit analog-to-digital converter when measuring voltages to show on the scope's screen?

5. When trying to analyze a circuit you should include everything in your schematic, and you shouldn't leave scope reference leads dangling. In this case Woopy's dangling reference clip is connected, through all the scope's probe cabling shields and chassis, to the same place the other probe's reference clip is connected. And the pulser circuit is also connected to both the mosfet Gate and to the circuit's negative rail (mosfet Source). At very low frequencies these facts may not matter much, but as you increase the operating frequency they will matter more and more, and sloppy technique should not be practiced or it will become habitual.


In short.... averaging before multiplying V and I does NOT yield a valid " average power " result, especially if you are using analog meters. And any result you get from any calculations, done by hand or by machine or by instrument, is subject to the errors in the input values. Know how these errors affect your results!  Components have tolerances, meters have offsets, ADCs have limits in precision. Know your instruments, and know how they actually work when calculating things like instantaneous multiplication of traces, signal average values, etc.

Let's use the correct terms and values and see what is actually happening here.

In the first place, the duty cycle is 20.9 percent, or 0.209, not 20 percent.
The CH1 measurement is not "Vcc", it is the V_drop across the 100 ohm (probably 5 or 10 percent tolerance) resistor.
The CH2 measurement is not "Vcc", it is the V_drop across the 1 ohm (probably 5 or 10 percent tolerance) resistor.

Let us "assume" that the 1 Ohm resistor is accurate, though.

So we have the voltage drop across the 1 ohm CVR as 134 mV, which translates to 134 mA.
Since this current is flowing through both resistors, we know the current through the 100R resistor is also 134 mA.
But the V_drop across that resistor is 12.6 volts. What does that tell you about the actual value of that resistor?

R=V/I, so 12.6/0.134 is 94.3 Ohms... just about almost within the 5 percent tolerance and well within the 10 percent tolerance of those _non-precision_ resistors.
But let's just let that slide for the moment.

The power dissipated by the 1 Ohm CVR during the ON time of the pulse, if the readings are accurate, is I²R = (0.134)x(0.134)x1 = 0.0180 Watt. But this is only occurring for 0.209 of the total cycle time, so the _average power_ dissipated by the CVR over the whole 1.0 time is 0.00375 Watt. This is the average power dissipated by the 1 ohm CVR, assuming it is accurately 1 ohm and the scope's voltage value is correct.

The power dissipated by the 94.3 ohm "100 ohm" load resistor, during the ON time of the pulse, is again I²R. We know the current is 0.134 A since we "trust" our one ohm CVR. So the power dissipated by the Load resistor is (0.134)x(0.134)x94.3 = 1.69 Watts. But this is only happening for 20.9 percent of the time (0.209). So the average power dissipated by the Load resistor is 1.69 x 0.209 = 0.353 Watt.

A quick "sanity check" shows that the 1 ohm CVR should be dissipating about 1/100 the power that the 100 ohm Load dissipates, since both are carrying the same current being in series. ("About" because of the tolerance ratings of the resistors. If the 1R is accurate and the scope's voltages are accurate, then the 100R is actually 94.3 Ohms.)


The "average current" is 0.134 A x .209 = 0.028 A, not 0.0204 A.

(ETA: Here the scope seems to be making some error. If we believe the "134mA" value and the "20.9" duty cycle value, the 28 mA value is correct, even though the scope is reporting "20.4 mA" as the mean for that channel. More probably, the "1 ohm" resistor is not actually one ohm, so the "134 mA" value is probably incorrect. But my calculations assume that it is exactly one ohm.)

Why are you multiplying the V_drop across the 100R by the "average current"?  What does this value mean? The power dissipated in the 100 R resistor is calculated by I²R, or equivalently V²/R.

This calculation you've made here is totally invalid.

Since we "know" that it is not correct to take the average values first and then multiply them, this calculation is invalid.

Here you are approximately correct, for the values you are using. The power dissipated in the 100R is
V²/R which indeed results in 1.587 Watts (ignoring the fact that the resistor is actually not 100R.)
The duty cycle is not 20 percent, it is nearly 21 percent, but your 317 mW is correct for the values you are using.

Using the correct values for the resistor (94.3 ohms) and the duty cycle (0.209) we get 1.683 Watts peak
and
(12.6)²/94.3 =  1.683 peak
1.683x0.209 = 0.351 Watts average. Which is in agreement with the values we got by doing Version A _correctly_.
Again, invalid because we do not multiply the means to get an average power. We multiply the actual values to get instantaneous power, then we find the average of _that_ to get average power.

You aren't doing it right !!!!!!!!

In the first place, we can completely reject the calculations that start with the average values of voltage and current. It is simply WRONG to try to get an average power value this way!

In the second place, the duty cycle is not 20 percent, it is 20.9 percent according to the scope.  This is a significant error you have introduced by using the wrong value for the duty cycle.

In the third place you are going badly wrong in some of your calculations.

The values from correct calculations based on the readings, and assuming that the one-ohm resistor is precisely accurate, are these (rounded to 3 sig digits):

The dutycycle (from the scope reading) is 20.9 percent.
The "100R" resistor is actually 94.3 ohms.
Average Power dissipated in the 1 ohm CVR is 0.00375 Watt.
Average Power dissipated in the 94.3 ohm Load resistor is 0.353 Watt.

Note that the power dissipated in the load resistor is 94 times the power dissipated in the CVR.... as it should be since both are carrying the same current and the Load resistor is 94 times the resistance of the CVR.

A 0.1 Ohm Current Sense Resistor would have a much lower Voltage drop Across the resistor and have a much lower over all affect on the circuits, if this helps? Also, why not use the CVR or Current Sense Resistor (CSR) on the Negative Rail? It might be easier to have the Scope Grounds closer to the Negative Terminal of the Power Source. Not saying its wrong, but I always think it is a best practice to minimize any possible Ground Loops.

There is always going to be some Inductance in the Current Sensing Circuit, but trying to minimize it as much as possible is also a best practice: 0.1 ohm, 5 W, ± 1%, Open Element, Through Hole Current Sense Resistor - Not a bad investment for $3

Again, you can use ohms law to calculate the Ampere Value flowing through the resistor: I = V/R - 0.05 / 0.1 = 0.5 Ampere's as an example.

TK Thanks again for such a detailed post pointing out the traps. Very nice to see such good info here for all!

   Chris Sykes
       hyiq.org